100 mL of 0.01 M XO_(4)^(-) is reduced to X^(n+) by 100 mL of 0.05 M Fe^(2+) in acidic medium. Thusoxidation state of X in X^(n+) is ______________

100 mL of 0.01 M XO_(4)^(-) is reduced to X^(n+) by 100 mL of 0.05 M F

1.	100 mL of 0.01 M XO_(4)^(-) is reduced to X^(n+) by 100 mL of 0.05 M Fe^(2+) in acidic medium. Thusoxidation state of X in X^(n+) is ______________
Answer» SOLUTION :`100xx0.05xx1=100xx0.01xx(7-n)n=2`